Sakai's Radon Nikodym Theorem for von Neumann algebra goes as follows:
Let $\phi$ and $\psi$ be normal forms on a $von$ $Neumann$ $algebra$ $M$ such that $\phi$ $\leq$ $\psi$. Then $\exists$ $a$ $\in$ $M$ uniquely determined by the following conditions:
$0$ $\leq$ $a$ $\leq 1$
$s(a)$ $\leq$ $s(\psi)$
- $\phi=L_aR_a\psi$
The proof of this version of the theorem can be found in Stratila-Zsido starting on Page 125 whose pdf version is available online via Google Search. I want to show that this is equivalent to the Radon Nikodym Theorem that one encounters in Measure Theory when $M = L^{\infty}(X,A,\mu)$ for some finite measure $\mu$ with the condition that the Radon Nikodym derivative is bounded by 1. Please help me with this. In this context, I shall be grateful if someone can also explain to me ( with proof ) the analogous notions of $s(a)$ and $s(\psi)$ when $M = L^{\infty}(X,A,\mu)$ for some finite measure $\mu$. I guess that they are somehow related to the support of measures. Some defintions of terms used above -
- Normal forms are $\sigma$ - weakly continuous positive forms.
- $\phi$ $\leq$ $\psi$ $\iff$ $\psi(x)$ - $\phi(x)$ $\geqslant$ $0$ $\forall$ $x$ $\geqslant$ $0$.
- $s(a)$ = projection onto the closure of the range of $a$.
- $1$ - $s(\psi)$ = $sup$ { $p$ $\in$ $P_M$ : $\psi(p) = 0$ } where $P_M$ denotes the set of all projections in $M$.
- $(R_a\psi)(x)$ = $\psi(xa)$ and $(L_a\psi)(x)$ = $\psi(ax)$ $\forall$ $x$ $\in$ $M$.
Thanks for any help.
Sakai's theorem does not imply Radon-Nikodym's theorem in its full generality. Because the $a$ Sakai's theorem gives is bounded, while in general the Radon-Nikodym derivative can be unbounded.
For instance if $\mu$ is Lebesgue measure on $[0,1]$, you can take $$ \nu(E)=\sum_{n=1}^\infty 2^n\,\mu\left(E\cap\left(\frac1{2^n},\frac1{2^{n-1}}\right]\right). $$ Then $\nu$ is absolutely continuous with respect to $\mu$, but the linear functional it produces on $L^\infty[0,1]$ is unbounded. In fact, the Radon-Nikodym derivative is the unbounded function $$ h=\sum_{n=1}^\infty2^n\,1_{\left(\frac1{2^n},\frac1{2^{n-1}}\right]}. $$
The converse also has problems: in general, not every state of $L^\infty(X)$ is given by integration against a measure (in the usual sense, at least).